Proof of the Double Bubble Conjecture in R^ N

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Figure 2. A nonstandard minimal double bubble must be a hypersurface of revolution about an axis L, consisting of a central bubble with layers of toroidal bands. Here we show the generating curves of a typical 4+4 double bubble, together with the associated tree T .

Figure 3. The generating curves for another possible 4+4 bubble with the same tree structure as in Figure 2.

In 2003, Reichardt, Heilmann, Lai and Spielman [RHLS03] extended their arguments to R4, and to higher dimensions Rn provided one volume is more than twice the other. In these cases, component bounds guarantee that one region is connected and the other has a finite number k of components. Eliminating 1 + k bubbles proved the conjecture in these cases. The same component bounds show that in R5 it suffices to eliminate 2 + 2 bubbles (as well as 1 + k bubbles) to prove the Double Bubble Conjecture. In R6 it suffices to eliminate also 2 + 3 bubbles. However, in Rn generally we know only that the larger region has at most three components and the smaller region has a finite number of components [HLRS99]. Here, we extend the methods of Hutchings et al. and Reichardt et al. to prove the Double Bubble Conjecture in Rn for n ≥ 3 for arbitrary volumes. We prove that j + k nonstandard bubbles are not minimizing for arbitrary finite component counts j, k. The arguments are similar in spirit to those of [RHLS03], except we take advantage of more properties of constant-mean-curvature surfaces of revolution in order to eliminate previously problematic cases. Our arguments also simplify the previous proofs in R3 and R4 because they eliminate the need for component bounds. 1.2. The Instability Argument. An area-minimizing double bubble Σ exists and has an axis of rotational symmetry L. Assume that Σ is a nonstandard double bubble. Consider small rotations about a line M orthogonal to L, chosen so that the n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 3

Figure 4. The lines orthogonal to Σ through the points of the separating set all pass through M. Σ cannot be a minimizer. points of tangency between Σ and the rotation vectorfield v separate the bubble into at least four pieces, as in Figure 4. Then we can linearly combine the restrictions of v to each piece to obtain a vectorfield that vanishes on one piece and preserves volume. By regularity for eigenfunctions, v is tangent to certain related parts of Σ, implying that they are spheres centered on L ∩ M. In turn, this implies that there are too many spherical pieces of Σ, leading to a contradiction. This is the instability argument of [HMRR02] behind Theorem 4.2. Therefore, no such useful perturbation axis M can exist. By induction, starting at the connected components corresponding to leaves in the tree of Figure 2 we classify all possible configurations in which no such M can be found. The induction ultimately shows that Σ must be a “near-graph component stack.” A global argument then finds a suitable M. Therefore, Σ cannot in fact be a minimizer. Having eliminated all nonstandard double bubbles from consideration, the only possible minimizer left is the standard double bubble.

2. Delaunay hypersurfaces Constant-mean-curvature hypersurfaces of revolution are known as Delaunay hypersurfaces [HMRR02, Hsi82, HY81, Del41, Eel87]. Let Σ ⊂ Rn be a constant- mean-curvature hypersurface invariant under the action of the group O(n) of isome- tries ﬁxing the axis L. Σ is generated by a curve Γ in a plane containing L. Put coordinates on the plane so L is the x axis. Parameterize Γ = {x(t),y(t)} by arc- length t and let θ(t) be the angle from the positive x-direction up to the tangent to Γ. Then Γ is determined by the diﬀerential equations x˙ = cos θ y˙ = sin θ (1) cos θ θ˙ = −(n − 1)H + (n − 2) y 4 BEN W. REICHARDT

Here, tan θ =y/ ˙ x˙ is the slope of Γ, κ = −θ˙ is the planar curvature of Γ with respect to the normal N = (sin θ, − cos θ), and H is the (constant) mean curvature of Σ with respect to N (an average of the planar curvature κ and the curvature due to rotating about L). The force of Γ with respect to N is a constant given by (2) F = yn−2(cos θ − Hy) . Note that from Eqs. (1) and (2), (n − 1)(n − 2) (3) θ¨ = − F sin θ . yn

Theorem 2.1 ([HMRR02, Prop. 4.3]). Let Γ be a complete upper half-planar generating curve that, when rotated about L, generates a hypersurface Σ with constant mean curvature. The pair (H, F ) determines Γ up to horizontal translation. (1) If H = 0 and F 6= 0, then Γ is a curve of catenary type and Σ is a hypersurface of catenoid type. (2) If HF < 0, then Γ is a locally convex curve and Σ is a nodoid. (3) If HF > 0, then Γ is a periodic graph over L and Σ is an unduloid or a cylinder. (4) If H = F =0, then Γ is a ray orthogonal to L and Σ is a vertical hyperplane. (5) If F =0 and H 6=0, then Γ is a semi-circle and Σ is a sphere. See Figure 5. The Delaunay hypersurfaces with nonzero mean curvature are the sphere, unduloid and nodoid. If Σ has positive mean curvature upward then it must be a nodoid. If Γ is not graph, then Σ must be either a nodoid or a hyperplane. Lemma 2.2 (Force balancing [HMRR02, Lemma 4.5]). Assume that three generating curves Γi, i =0, 1, 2, of Delaunay hypersurfaces meet at a point p. Consider normals turning clockwise about p. If the curvatures with respect to these normals satisfy H0 + H1 + H2 =0, then the forces with respect to these normals satisfy

(4) F0 + F1 + F2 =0 . The lemma follows from Eq. (2).

3. Structure of minimal double bubbles A double bubble is a piecewise-smooth oriented hypersurface Σ ⊂ Rn consisting of three compact pieces Σ1, Σ2 and Σ0 (smooth on their interiors), with a common boundary such that Σ1 ∪ Σ0, Σ2 ∪ Σ0 enclose two regions R1, R2, respectively, of given volumes. The work of Almgren [Alm76] (see [Mor00, Ch. 13]) and Hutchings establishes the existence and structure. Theorem 3.1 ([Hut97, Theorem 5.1]). Any nonstandard minimal double bubble is a hypersurface of revolution about some line L, composed of pieces of constant- mean-curvature hypersurfaces meeting in threes at 120 degree angles. The bubble is a topological sphere with a ﬁnite tree T of annular bands attached, as in Figure 2. The two caps of the bottom component are pieces of spheres, and the root of the tree has just one branch. Hence, any minimal double bubble is determined by an upper half-planar diagram of generating curves that, when rotated about L, generate the double bubble. n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 5

Figure 5. Smooth regions of the cluster are parts of “Delaunay” hypersurfaces of revolution. Plotted are the generating curves of Delaunay hypersurfaces in R3 all starting at (x, y) = (0, 1) and θ = 0, with increasing mean curvature from top to bottom (positive downward at x = 0): three unduloids, a sphere, then three nodoids. Below are examples of a catenoid and a vertical hyperplane, the Delaunay hypersurfaces of zero mean curvature.

By studying these generating curves, we will eliminate as unstable nonstandard double bubbles. If a double bubble is stable, then for i =1, 2, the mean curvature Hi of Σi with respect to normals pointing into the region Ri is constant over the various smooth pieces of Σi and is called the pressure of Ri, positive by [Hut97, Corollary 3.3]. 6 BEN W. REICHARDT

The pressure diﬀerence H0 = H1 − H2 is the constant mean curvature of Σ0 with respect to normals pointing from R2 into R1 [HMRR02, Lemma 3.1]. Therefore Lemma 2.2 applies to each vertex, implying. Claim 3.2. The outer boundaries of a connected component C – between C and n the outside, R \ (R1 ∪ R2) – are parts of the same Delaunay hypersurface up to horizontal translation. Proof. The boundary pieces have the same pressure H and – by applying Lemma 2.2 repeatedly – the same force F , so are parts of the same Delaunay hypersurface up to horizontal translation by Theorem 2.1. More carefully, the proof is by induction on the height of the subtree of descen- dants of C in the associated tree T (from Theorem 3.1). Figure 3 provides examples of the cases that arise. • The base case is when C has no children, so has just a single piece of outer boundary Γ (e.g., Γ5 in Figure 3). Then the claim is trivial. • To step the height inductive assumption, assume, e.g., that Γ3 and Γ7 have the same force. Force-balancing Lemma 2.2 on vertices v123 ≡ Γ¯1 ∩ Γ¯2 ∩ Γ¯3 and v278 implies that Γ2 and Γ8 have the same force with respect to normals pointing into C. The same arguments show that Γ9 and Γ10 – so all outer boundaries of C – have the same force.

Although we will not need it, it is also interesting to remark: Lemma 3.3 ([HMRR02, Lemma 6.4]). In a minimizing double bubble for unequal volumes, the smaller region has larger pressure. Lemma 3.3 follows from concavity of the minimum-area function [Hut97, Theo- rem 3.2]. Note however that in a minimizer for equal volumes, the regions may still have unequal pressures; H0 need not be zero.

4. Instability by separation Let Σ ⊂ Rn be a regular stationary double bubble of revolution about axis L, with upper half planar generating curves ∪Γ¯i consisting of arcs Γ¯i, with interiors Γi, ending either at the axis or in threes at vertices vijk = Γ¯i ∩ Γ¯j ∩ Γ¯k.

4.1. The map l : ∪Γi → L ∪ {∞}.

Definition 4.1. Define a map l : ∪Γi → L ∪{∞} ≡ [−∞, +∞]/(−∞ ∼ +∞) that maps p to the intersection with L of the line orthogonal to Γ at p. Let L(p) be the −−→ ray pl(p). Let lΓ be the restriction of l to the arc Γ. For p an endpoint of Γ, define lΓ(p) ∈ [−∞, +∞] to be the limiting value of lΓ approaching p, limq→p l(q). (For Γ q∈Γ a vertical hyperplane, lΓ(p)= ∞.)

For short, we will write li for lΓi , and use similar abbreviated notation for other relevant variables L, θ, κ, N, F . For p an endpoint of Γi, note that if li(p) ∈ l(Γj ) and Γj is not a circle or hyperplane (Fj 6= 0), then for all q ∈ Γi suﬃciently close to p, l(q) ∈ l(Γj). (This follows, e.g., because l˙j is proportional to the force Fj ; see [HMRR02, Remark 5.1].) n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 7

4.2. Separating set l−1(x) instability claims. Theorem 4.2 ([HMRR02, Prop. 5.2]). Consider a stable double bubble of revolution Σ ⊂ Rn, n ≥ 3, with axis L. Assume that there is a minimal set of points {p1,...,pk} in ∪Γi with l(p1)= ··· = l(pk)= x that separates ∪Γ¯i. Then every connected component of Σ that contains one of the points pi is part of a sphere centered at x (if x ∈ L) or part of a hyperplane orthogonal to L (in the case x = ∞). We sketched the proof of Theorem 4.2 in our introduction Section 1.2. We will apply several useful corollaries of this theorem, taken directly from [RHLS03]. Corollary 4.3 ([RHLS03, Cor. 4.2]). No generating curve that turns downward past the vertical can have an internal separating set, i.e., two points p1 6= p2 in the arc, with l(p1)= l(p2). Proof. Otherwise, by Theorem 4.2, the arc would have to be part of either a circle with center on the axis L or a line perpendicular to L. But neither turns past the vertical. Corollary 4.4 ([RHLS03, Cor. 4.3]). No generating curve that is not part of a vertical line can go vertical twice, including at least once in its interior. Proof. Such an arc (a nodoid by Theorem 2.1) has a separating set l−1(x) for some x with |x| large enough, contradicting Corollary 4.3. Corollary 4.5 ([RHLS03, Cor. 4.4]). Consider a nonstandard minimizing double bubble. Then there is no x ∈ L∪{∞} such that l−1(x)r(two circular caps) contains points in the interiors of distinct Γi that separate ∪Γ¯i. Proof sketch. For x ∈ L, the statement is [HMRR02, Prop. 5.7]. Arguments using “force balancing” show that more pieces of the minimizer are spherical and hence the bubble is the standard double bubble. For x = ∞, note that a separating set crosses at least one outer boundary. By Theorem 4.2, this boundary is a vertical line, contradicting positive pressure of the regions. We will consider various nonstandard double bubbles, and show that they violate one of the above corollaries of Theorem 4.2, hence cannot be minimizing. These corollaries suﬃced for the proof of the Double Bubble Conjecture in R4, but in higher dimensions Rn we will need to use more information about Delaunay hypersurfaces, Lemmas 5.1 and 5.2.

5. Delaunay hypersurface lemmas We will need a few more properties of Delaunay hypersurfaces, which can be proven by comparing Σ to certain spheres. Lemma 5.1. Consider an unduloid generated by Γ rising to a point p where it makes an angle θΓ(p) ∈ [π/2 − 2ψ,π/2) above the horizontal. Then the ray going back down from p at an angle of ψ ∈ (0,π/4) with the unduloid passes completely beneath it to the left of p.

Proof. For θΓ(p) ≥ π/2 − ψ, the ray leaves p heading to the right, so cannot cross the unduloid to the left. Assume therefore that θΓ(p) ∈ [π/2 − 2ψ,π/2 − ψ]. (In applications, we will set ψ = π/6.) See Figure 6. 8 BEN W. REICHARDT

Figure 6. Lemma 5.1.

Figure 7. The ray from p stays within the upper half of the circle C centered on the x axis and tangent to Γ at p. By comparing curvatures, C stays below Γ to the left of p.

Consider the circle C centered on the x axis and passing through p tangent to the unduloid. Note that the ray in question passes entirely beneath the semicircle above the x axis; Figure 7. We will show that the unduloid stays above the semicircle to the left of p. Recall Eq. (2) for the force F = yn−2(cos θ − Hy) of a constant-mean-curvature surface of revolution with respect to a downward-pointing normal. For an unduloid F > 0, while for a sphere F = 0 (Theorem 2.1). Since at p, y and θ are the same, the sphere has higher mean-curvature H. In particular, κΓ(p)= −θ˙Γ(p) the planar curvature of Γ at p is less than κC (p) (because the portion of H due to rotating about L is the same for each), so Γ leaves p above C. Assume for contradiction that Γ crosses C for the first time at a point q to the left of p. Then the unduloid is steeper than the circle at q: θΓ(q) >θC (q), so C achieves the angle θΓ(q) at some point below q. However, this is impossible; for any angle φ, the y-coordinate at which Γ first reaches φ, y(p) − ∆yΓ(φ), is no more than the y-coordinate at which C reaches φ, y(p) − ∆yC (φ). Indeed, θ¨Γ < 0 (Eq. (3)), so κΓ is decreasing moving left, while κC is constant. Γ therefore reaches some maximum angle φmax ≥ θΓ(p). Beyond φmax, Γ turns upward away from the circle, so the curves can only further diverge. But for φ ∈ [θΓ(p), φmax], the difference in y coordinates from p to where the curve reaches φ is φ dt ∆y = dψ sin ψ ZθΓ(p) dψ n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 9

φ 1 = dψ sin ψ ZθΓ(p) κ(ψ)

Since κΓ(ψ) is decreasing, while κC is constant, indeed ∆yΓ ≥ ∆yC . Lemma 5.2. Consider a smooth curve Γ of length ∆t between points p and q. Let θ(t) be the angle the curve makes with the horizontal at arc-length t from p, with θp = θ(0) and θq = θ(∆t) the initial and ﬁnal angles. Assume θp ≤ θq ≤ θp + π/3. (If we rotate the coordinate system so θp = 0, then 0 ≤ θq ≤ π/3.) Assume θ¨ < 0 and θ(t) in [θp,θq]. Then the line through q at angle π/6 clockwise from Γ (i.e., at angle θq − π/6 above horizontal) passes above Γ.

Proof. Rotate the coordinate system so without loss of generality θp = 0 and θq ≤ π/3. Then θ<¨ 0 implies that for all t ∈ [0, ∆t], t θ(t) > ∆t θq . Let ∆y = y(∆t) − y(0) ∆t = sin(θ(t))dt Z0 ∆t t > sin( ∆t θq)dt Z0 = ∆t(1 − cos θq)/θq , where for the inequality we used that sin φ is increasing for φ ∈ (−π/2,π/2). Sim- ilarly, let ∆x = x(∆t) − x(0) ∆t = cos(θ(t))dt Z0 ∆t t < cos( ∆t θq)dt Z0 = ∆t(sin θq)/θq , using that cos φ is decreasing for φ ∈ (0, π). Hence, ∆y 1 − cos(θ ) > q ∆x sin(θq)

≥ tan(θq − π/6) . Γ is a convex-up curve between p and q, so it lies beneath the direct line segment pq, which has higher slope than, and hence lies beneath, the line through q at angle θq − π/6 above horizontal. By Theorem 2.1 and Eq. (3), on a clockwise-turning nodoid, θ¨ < 0 provided y˙ = sin θ < 0; while for an unduloid with t increasing to the right, θ¨ < 0 provided y>˙ 0. Corollary 5.3. Consider an unduloid or nodoid generating curve strictly increasing left to right from p to q with the notation of Figure 8. Assume θ3(p) ∈ [0,π/2) and

max{π/6,θ3(p)} <θ3(q) ≤ min{π/2,θ3(p)+ π/3} . 10 BEN W. REICHARDT

Figure 8. Notation for a boundary involving ﬁve arcs, Γ1 through Γ5, for Corollaries 5.3 and 5.4.

Then l4(q)

−−−→ Proof. By Lemma 5.2, the ray L4(q)= ql4(q) (the extension of the normal N4(q)) passes above p and by assumption it is clockwise from N2(p), so it stays above −−−→ L2(p)= pl2(p). Corollary 5.4. Consider a piece of boundary, arriving at q from the left with the notation of Figure 8. Assume θ3(q) ∈ [π/6,π/2) and θ3(p) ∈ [π/6,θ3(q)+ π/3]. Also assume that θ3 6= π mod 2π between p and q. Then l5(q) >l1(p).

Proof. There are three cases, depending on whether Γ3 is an unduloid, a convex-up catenoid or nodoid, or a nodoid convex-down at q. The assumption that θΓ3 6= π mod 2π between p and q prevents a nodoid from turning too far around between p and q.

(1) If Γ3 is an unduloid, then by Lemma 5.1, the ray L5(q) passes downward beneath Γ3, so in particular stays right of p. (This remains true if the unduloid passes through one or more minima between p and q.) (2) If Γ3 is a convex-up catenoid or nodoid, then convexity implies again that L5(q) stays right of Γ3. (3) IfΓ3 is a convex-down nodoid, then by Lemma 5.2 with ψ = π/6 again L5(q) stays right of Γ3. (Applying the lemma requires reﬂecting horizontally and using that Γ3 turns no more than π/3 radians from p to q.)

In all cases, since Γ1 leaves p clockwise of how Γ5 leaves q, l1(p)

6. Rotation notation A nonstandard minimal double bubble’s generating curves can be classiﬁed by the angles at which arcs leave each vertex. For our analysis, it will suﬃce to know which arcs are leaving from the right and which from the left; there are therefore six cases, each covering π/3 radians, which we term “notches.” Incrementing the rotation notch of a vertex corresponds to turning it counterclockwise until an arc passes the vertical, as occurs for Γ3 from Figure 3 to Figure 9, and from Figure 10(a) to 10(b). The extreme position with an arc leaving a vertex exactly at the vertical divides two consecutive notch cases. If the limiting value of l along the vertical arc is +∞ (or if the arc is a vertical line), the position is assigned the smaller rotation n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 11

Figure 9. From the curves of Figure 3, vertex p = Γ¯1 ∩ Γ¯2 ∩ Γ¯3 has turned one counterclockwise “notch,” since Γ3 has passed the vertical.

Figure 10. A close-up view of p as it turns one notch counterclockwise. In (a), l2(p)

7. Near-graph leaves and bubble stacks 7.1. Near-graph leaves. A “leaf” of a nonstandard minimizing double bubble corresponds to a leaf of its associated tree of Theorem 3.1 and Figure 2. We will use the notation of Figure 12. Deﬁnition 7.1. A leaf is near graph if vertices p and q are rotated respectively (0, 0) notches, (0, 1) notches or (−1, 0) notches from their positions in Figure 12. (Figure 13 gives examples for the latter two cases.) We say that a near-graph leaf is right-side up if, with the notation of Figure 12, the region below Γ2 is inside the bubble; and upside down if the region above Γ3 is inside the bubble. 12 BEN W. REICHARDT

Figure 11. This nonstandard double bubble has the same associated tree structure as those in Figures 3 and 9, but some vertices have been rotated a notch or two counterclockwise or clockwise, as marked.

Figure 12. A leaf involves four arcs: Γ1, Γ2, Γ3, Γ4. In general each vertex can be rotated m notches counterclockwise from the pictured conﬁguration, in which all arcs are graphs and mp = mq = 0.

Figure 13. Leaves with p and q vertices rotated respectively (0, 1) notches counterclockwise from their positions in Figure 12 (left), or (−1, 0) notches (right).

Claim 7.2. For an upside-down, near-graph leaf component, the region containing the leaf must have strictly higher pressure.

Proof. If Γ3 has positive pressure up in Figures 12 or 13, then Γ3 must be a convex- up nodoid. But then the angle constraints at vertices p and q cannot be satisﬁed, a contradiction. If there is zero pressure across Γ3, then Γ3 can be a catenoid or a vertical hyperplane, but again neither matches the angle constraints. 7.2. Right-side-up, near-graph component stacks. Deﬁnition 7.3. A component stack consists of a base component and all its descen- dants in the associated tree T . The base component cannot be the root component of T . A right-side-up, graph component stack is a component stack in which all the n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 13

Figure 14. Two examples of right-side-up, near-graph component stacks. The ﬁrst is in fact a graph stack.

Figure 15. Notation for right-side-up, near-graph component stacks. associated generating curves are graph, and the root is downward from the base component. A right-side-up, near-graph component stack – or “near-graph stack,” for short – is the same, except for each internal boundary the left vertex can be rotated one notch clockwise or the right vertex rotated one notch counterclockwise.

Examples are in Figures 12 and 13 (if the path to the root passes through Γ2), and in Figure 14. Lemma 7.4. For a right-side-up, near-graph component stack in a minimizer, with the notation as in Figure 15, l1(p) l1(p) and l3(q)

Figure 16. Notation for the base component of a stack of bubbles. On the left (right, respectively), the root of the associated tree is downward (upward, resp.), and leaves are upward (downward).

If the component stack has height greater than one, then by induction and to prevent internal separating sets l(Γ3) < l(Γ4) < ···

Proof. By Lemma 7.4, l1(p)

8. Leaf and component classification We now classify the types of leaves and general components that may occur in a minimizer. In flavor, this section is similar to the leaf classification [RHLS03, Prop. 7.1] – however, we will complicate things by considering component stacks and not just leaves; and simplify things by applying Corollaries 5.3 and 5.4. Theorem 8.1. In a minimizer, every leaf must be near graph, and every component stack of height > 1 must be near graph and right-side up. Proof. The proof will be by induction on the height of the stack. To reduce repet- itive arguments, we will simultaneously consider leaves and component stacks; on a first reading, however, it may be easier to specialize just to leaves. Diagrams will be condensed by using to indicate the possible presence of one or more near-graph leaves or right-side-up, near-graph stacks. Notation: For leaves we will use use the notation of Figure 12. Cases will be separated according to vertex rotations in notches away from their positions in that diagram. For the base component of a component stack of height > 1, cases will be separated according to vertices’ notches away from their positions in one of the cases of Figure 16. Here, either Γ2 or Γ3 is no longer a single arc, but the union of multiple outer boundary arcs (all pieces of the same Delaunay hypersurface by Claim 3.2). Note also the symmetries; case (i, j) is symmetrical under relabeling to (j −3,i− 3), and under horizontal reflection to (−j, −i). The most difficult cases will be (0, 2) (symmetrical to (−1, −3)) and (−1, −2). • Cases (0, 0) and (0, 1): See Figures 17 and 18(a). This case includes all leaves or base components with θ1(p) ∈ [−π/6,π/6] and θ4(q) ∈ [−π/6,π/2) (and θ4(q) = π/2 if l4(q)=+∞). A (0, 0) or (0, 1) leaf is near graph, so is allowed in the n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 15

Figure 17. Case (0,0). Either a (0, 0) leaf – i.e., in which vertices p and q are each rotated zero notches from their positions in Figure 12 – or the (0, 0) base component of a stack. In the latter case, the upper indicates the possible presence of one or more right-side-up, near-graph component stacks along Γ3, while the lower indicates the possible presence of upside-down, near- graph leaves along Γ2. statement of the theorem. If instead the component is the base of a component stack with one or more near-graph stacks sitting above it, overall we still have an allowed near-graph stack. If it has a near-graph leaf sitting below it, then by Claim 7.2 the leaf’s region has higher pressure. However, Γ3 must have positive pressure down to match the angle constraints at p and q, a contradiction. • Case (0, 2): See Figure 18(b). (With the notation of Section 6, this includes the case in which Γ3 is a vertical hyperplane.) A right-side-up, near-graph stack cannot be placed along Γ2 without contradicting Corollary 7.5, since Γ2 turns downward past the vertical. The presence of upside-down, near-graph leaves sitting beneath Γ2 beyond where Γ2 turns past the vertical will not aﬀect our argument. If Γ3 is a single piece of arc, then by Corollary 5.4 on Γ3, l2(p) < l2(q). (If Γ3 is a vertical hyperplane, then Corollary 5.4 doesn’t apply, but still l2(p) < l2(q).) However, [−∞,l2(q)) ⊂ l(Γ2), so Γ2 has an internal separating set, contradicting Corollary 4.3 or Corollary 4.5 (if there is a leaf below Γ2).

In Figure 18(b), Γ3 is drawn with θΓ3 ∈ [π/6,π/2] between p and q, so it may appear that a near-graph stack cannot be placed on Γ3. However, this is not the case; see, e.g., Figure 19. We claim that even if there are one or more near-graph stacks along Γ3, still l2(p) l2(p), so in particular l2(q) >l2(p) as claimed.

• Case (0, 3) or mq ≥ 3: See Figure 18(c). There cannot be a stack sitting above Γ2 without contradicting Corollary 7.5, since Γ2 turns downward past the vertical. 16 BEN W. REICHARDT

(a) Case (0,1) (b) Case (0,2) (c) Case (0,3)

(d) Case (0,-1) (e) Case (0,-2) (f) Case (-1,-2)

(g) Case (-1,-1) (h) Case (-1,1) (i) Case (1,-1)

Figure 18. Diﬀerent (mp,mq) component cases we consider. See also Figure 17. Case (i, j) is symmetrical under relabeling to (j − 3,i − 3), and under horizontal reﬂection to (−j, −i).

Figure 19. Left: A right-side-up, near-graph component stack might be placed on Γ3 in Figure 18(b) if Γ3 is an unduloid. Right: We show that nodoid Γ has an internal separating set by horizon- 2 −−−→ tally translating the ray L2(q)= ql2(q) to the piece of arc leaving p, then applying Lemma 5.1. n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 17

Figure 20. In cases (0, −1) and (0, −2), if Γ2 is a piece of unduloid, and goes through a minimum and maximum between p and q, then there is a point r ∈ Γ2 at the same height as p with l(r) >l1(p). Therefore l1(p) ∈ l(Γ2).

Γ3 must be a convex-left nodoid (it cannot have any components along it by in- −−−→ duction), so the ray L2(q)= ql2(q) stays right of Γ3, implying l2(p) ∈ [−∞,l2(q)) ⊂ l(Γ2), contradicting either Corollary 4.3 or Corollary 4.5 (if there is a leaf below Γ2). • Cases (0, −1), (0, −2) and (−1, −2): See Figures 18(d), 18(e) and 18(f). There cannot be a right-side-up near-graph stack above Γ3 by Corollary 7.5 since Γ3 turns downward past the vertical. Upside-down, near-graph leaves cannot be placed along Γ2 without contradicting Claim 7.2. The presence of upside-down, near-graph leaves sitting beneath Γ3 beyond where Γ3 turns past the vertical will not aﬀect our argument. Now l1(p) ∈ [−∞,l3(p)) ⊂ l(Γ3). In case (0, −1) or (0, −2), l2(p) y(q)), we get l3(p) − l1(p) >l3(q) − l2(q). Since l2(q) ≤ l1(p), l3(q)

1[HMRR02, Lemma 5.9] would suﬃce for this case, and in fact with Corollary 5.3 is enough for all of cases (0, −1) and (0, −2). However, we will give a slightly simpler argument just for case (0, −1) with θ2(q) ≥ 0. 18 BEN W. REICHARDT

Figure 21. Lemma 8.2.

• Case (−1, −1): See Figure 18(g). There cannot be a right-side-up near-graph stack above Γ3 by Corollary 7.5, nor an upside-down near-graph leaf under Γ2 by Claim 7.2. It remains to eliminate the case of a (−1, −1) leaf – we repeat the argument of [RHLS03, Prop. 7.1]. Now l1(p) ∈ [−∞,l3(p)) ⊂ l(Γ3). The ray L1(p) stays left of the convex-right nodoid Γ2, so l1(p)

• Case (−1, 1) or mq ≥ 1: See Figure 18(h). Γ2 goes twice vertical, contradicting Corollary 4.4 or Corollary 4.5.

This and symmetrical considerations concludes the argument if either Γ1 or Γ4 descends approaching p or q, respectively. The last remaining case is (1, −1), in which both Γ1 and Γ4 ascend approaching p or q.

• Case (1, −1) or mq ≤−1: See Figure 18(i). Γ3 goes twice vertical, contradicting Corollary 4.4 or Corollary 4.5.

π Lemma 8.2. Let r be a point at height y(r) > 0 above the axis L. Let φ ∈ [0, 6 ). π Drop lines from r at angles φ and φ + 3 from the vertical (Figure 21). Then the diﬀerence in the x-coordinates of these lines’ intersections with L, π y(r) tan(φ + 3 ) − tan φ , is a strictly increasing function in both φ and y(r). Proof. Simplifying, π π tan 3 tan(φ + 3 ) − tan φ = 1 π , 2 + cos(2φ + 3 ) π 1 π and cos(2φ + 3 ) > 2 and is decreasing for φ ∈ [0, 6 ).

9. Root stability The “root” of a nonstandard minimizing double bubble corresponds to the root of its associated tree of Theorem 3.1 and Figure 2. The root involves ﬁve arcs including two circular caps to either side, as in Figure 22. Proposition 9.1. In a minimizer, the child component C of the root component cannot be either an upside-down, near-graph leaf, or the base component of a right- side-up, near-graph stack. n PROOF OF THE DOUBLE BUBBLE CONJECTURE IN R 19

Figure 22. The root component involves ﬁve arcs: Γ2, Γ3, Γ4, and the two circular caps Γ1 and Γ5.

Proof. Γ1 and Γ5 are pieces of circles centered on the axis L, so mp, mq the number of notches p, q are rotated from their positions in Figure 22 must each lie in {−1, 0, 1}. In particular, neither Γ1 nor Γ5 can turn past the vertical to connect to an upside-down, near-graph leaf. For C to be the base component of a right-side-up, near-graph stack, (mp,mq) ∈ {(0, 0), (−1, 0), (0, 1)}. In case (−1, 0), l4(q) < l3(q) and [−∞,l3(q)) ⊂ l(Γ3), so there is a separating set across Γ3, Γ4, contradicting Corollary 4.5. Similarly, (mp,mq) 6= (0, 1). In case (mp,mq) = (0, 0), to avoid a separating set between Γ3 and either Γ2 or Γ4, it must be that l(Γ4) < l(Γ3) < l(Γ2), contradicting Lemma 7.4 for the far endpoints of Γ2 and Γ4.

10. Proof of the Double Bubble Conjecture Proof of Theorem 1.1. Suppose that the minimizer is nonstandard. Each region has a ﬁnite number of components by Corollary 3.1. Consider the minimizer’s generating curves. Theorem 8.1 implies that the root’s child component is either an upside-down, near-graph leaf, or the base component of a right-side-up, near-graph component stack – contradicting Proposition 9.1. Therefore, an area-minimizing double bubble must be the standard double bubble.

Acknowledgements. The author thanks Marilyn Daily for helpful discussions, and thanks Frank Morgan for comments on an early draft of the paper. Research supported by NSF Grant PHY-0456720 and ARO Grant W911NF-05-1-0294.

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